如何查找Android设备的序列号？

正如 @haserman 所说：

TelephonyManager tManager = (TelephonyManager)myActivity.getSystemService(Context.TELEPHONY_SERVICE);
String uid = tManager.getDeviceId();

但是在清单文件中包含权限是必要的：

<uses-permission android:name="android.permission.READ_PHONE_STATE"/>

Build.SERIAL 是最简单的方法，但并不完全可靠，因为它可能为空，有时会返回与您设备设置中看到的不同的值(证明 1, 证明 2)。

根据设备制造商和Android版本，有几种获取该号码的方法，因此我决定将我能找到的每个可能的解决方案编译成一个单一的gist。以下是其简化版：

public static String getSerialNumber() {
    String serialNumber;

    try {
        Class<?> c = Class.forName("android.os.SystemProperties");
        Method get = c.getMethod("get", String.class);

        serialNumber = (String) get.invoke(c, "gsm.sn1");
        if (serialNumber.equals(""))
            serialNumber = (String) get.invoke(c, "ril.serialnumber");
        if (serialNumber.equals(""))
            serialNumber = (String) get.invoke(c, "ro.serialno");
        if (serialNumber.equals(""))
            serialNumber = (String) get.invoke(c, "sys.serialnumber");
        if (serialNumber.equals(""))
            serialNumber = Build.SERIAL;

        // If none of the methods above worked
        if (serialNumber.equals(""))
            serialNumber = null;
    } catch (Exception e) {
        e.printStackTrace();
        serialNumber = null;
    }

    return serialNumber;
}

Android操作系统设备的唯一设备ID，以字符串形式表示。

String deviceId;
    final TelephonyManager mTelephony = (TelephonyManager) getSystemService(Context.TELEPHONY_SERVICE);
        if (mTelephony.getDeviceId() != null){
            deviceId = mTelephony.getDeviceId(); 
         }
        else{
            deviceId = Secure.getString(getApplicationContext().getContentResolver(),   Secure.ANDROID_ID); 
         }

但我强烈建议使用谷歌推荐的这种方法:

识别应用程序安装

public static String getManufacturerSerialNumber() {
  String serial = null; 
  try {
      Class<?> c = Class.forName("android.os.SystemProperties");
      Method get = c.getMethod("get", String.class, String.class);
      serial = (String) get.invoke(c, "ril.serialnumber", "unknown");
  } catch (Exception ignored) {}
  return serial;
}

适用于API 29和30，在三星Galaxy S7、S9 Xcover上进行了测试

我知道这个问题很久了，但是它可以在一行代码中完成

String deviceID = Build.SERIAL;

从Android 10开始，应用程序必须具有READ_PRIVILEGED_PHONE_STATE特权权限才能访问设备的不可重置标识符，包括IMEI和序列号。

受影响的方法包括以下内容：

Build getSerial() TelephonyManager getImei() getDeviceId() getMeid() getSimSerialNumber() getSubscriberId()

READ_PRIVILEGED_PHONE_STATE仅适用于平台

我发现@emmby上面发布的示例类是一个很好的起点。但是它有一些缺陷，正如其他帖子中提到的那样。最主要的问题是它不必要地将UUID持久化到XML文件中，然后总是从该文件中检索它。这使得该类容易被黑客攻击：任何拥有root手机的人都可以编辑XML文件以获得新的UUID。

我已经更新了代码，只有在绝对必要时（即使用随机生成的UUID时）才将其持久化到XML，并按照@Brill Pappin的答案重新设计了逻辑：

import android.content.Context;
import android.content.SharedPreferences;
import android.provider.Settings.Secure;
import android.telephony.TelephonyManager;

import java.io.UnsupportedEncodingException;
import java.util.UUID;

public class DeviceUuidFactory {
    protected static final String PREFS_FILE = "device_id.xml";
    protected static final String PREFS_DEVICE_ID = "device_id";

    protected static UUID uuid;

    public DeviceUuidFactory(Context context) {

        if( uuid ==null ) {
            synchronized (DeviceUuidFactory.class) {
                if( uuid == null) {
                    final SharedPreferences prefs = context.getSharedPreferences( PREFS_FILE, 0);
                    final String id = prefs.getString(PREFS_DEVICE_ID, null );

                    if (id != null) {
                        // Use the ids previously computed and stored in the prefs file
                        uuid = UUID.fromString(id);

                    } else {

                        final String androidId = Secure.getString(context.getContentResolver(), Secure.ANDROID_ID);

                        // Use the Android ID unless it's broken, in which case fallback on deviceId,
                        // unless it's not available, then fallback on a random number which we store
                        // to a prefs file
                        try {
                             if ( "9774d56d682e549c".equals(androidId) || (androidId == null) ) {
                                final String deviceId = ((TelephonyManager) context.getSystemService( Context.TELEPHONY_SERVICE )).getDeviceId();

                                if (deviceId != null)
                                {
                                    uuid = UUID.nameUUIDFromBytes(deviceId.getBytes("utf8"));
                                }
                                else
                                {
                                    uuid = UUID.randomUUID();

                                    // Write the value out to the prefs file so it persists
                                    prefs.edit().putString(PREFS_DEVICE_ID, uuid.toString() ).commit();
                                }
                            }
                            else
                            {
                                uuid = UUID.nameUUIDFromBytes(androidId.getBytes("utf8"));
                            } 
                        } catch (UnsupportedEncodingException e) {
                            throw new RuntimeException(e);
                        }



                    }

                }
            }
        }

    }


    /**
     * Returns a unique UUID for the current android device.  As with all UUIDs, this unique ID is "very highly likely"
     * to be unique across all Android devices.  Much more so than ANDROID_ID is.
     *
     * The UUID is generated by using ANDROID_ID as the base key if appropriate, falling back on
     * TelephonyManager.getDeviceID() if ANDROID_ID is known to be incorrect, and finally falling back
     * on a random UUID that's persisted to SharedPreferences if getDeviceID() does not return a
     * usable value.
     *
     * In some rare circumstances, this ID may change.  In particular, if the device is factory reset a new device ID
     * may be generated.  In addition, if a user upgrades their phone from certain buggy implementations of Android 2.2
     * to a newer, non-buggy version of Android, the device ID may change.  Or, if a user uninstalls your app on
     * a device that has neither a proper Android ID nor a Device ID, this ID may change on reinstallation.
     *
     * Note that if the code falls back on using TelephonyManager.getDeviceId(), the resulting ID will NOT
     * change after a factory reset.  Something to be aware of.
     *
     * Works around a bug in Android 2.2 for many devices when using ANDROID_ID directly.
     *
     * @see http://code.google.com/p/android/issues/detail?id=10603
     *
     * @return a UUID that may be used to uniquely identify your device for most purposes.
     */
    public UUID getDeviceUuid() {
        return uuid;
    }

是的。它是设备硬件序列号，而且是唯一的。因此，在API级别2.3及以上，您可以使用android.os.Build.ANDROID_ID来获取它。对于低于2.3 API级别的设备，请使用TelephonyManager.getDeviceID()。

您可以阅读此http://android-developers.blogspot.in/2011/03/identifying-app-installations.html。